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Is spontaneous symmetry breaking robust against weak perturbations to the Hamiltonian?
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Setup
Suppose we have some Hamiltonian $H$ which is known to exhibit spontaneous symmetry breaking (SSB), at least in some parameter regime. For simplicity, we might consider the 2D Ising model below its critical temperature.
Now suppose we add some weak perturbation such that the new Hamiltonian is
$$
H^prime = H + lambda W,
$$
where we assume that $W$ is bounded above by $H$, $lVert W rVert leq lVert H rVert$, and $lambda ll 1$ is a dimensionless parameter controlling the strength of the perturbation. Note that I do not assume that $W$ preserves the symmetries of $H$.
Question
Can we say anything about whether or not $H^prime$ will also exhibit SSB? Furthermore, if it does exhibit SSB, how are the new symmetry breaking states related to the old ones? Are they the same, at least up to some $mathcalO(lambda)$ error? If there is a phase transition, are the critical parameters the same?
I believe the folklore answer to this question is "$H^prime$ will still exhibit SSB for $lambda ll 1$, and the new symmetry breaking states are the same as the old ones", at least in the simple case of the Ising model. However, I would be interested to hear if any of this has been rigorously proven, again even if only for some specific models.
Edit
I will attempt to clarify my question. I am mainly interested in the equilibrium states of the perturbed Hamiltonian, and whether they break the symmetry of the original Hamiltonian, while still forming a closed space under the action of the original symmetry group.
Copying from my earlier comments, I believe one example of where this happens is with discrete time crystals. For concreteness let us consider a many-body localized time crystal, following this paper as a reference. There one drives a system of qubits periodically with a two-pulse protocol. This can be described by the Floquet operator
$$
U_f = exp(-i t_0 H_mathrmMBL) expleft(i t_1 sum_i sigma_i^xright),
$$
where $H_mathrmMBL$ is given by
$$
H_mathrmMBL = sum_i (J_i sigma_i^zsigma_i+1^z + h_i^z sigma_i^z + h_i^x sigma_i^x),
$$
with the parameters $J_i$, $h_i^z$ and $h_i^x$ chosen uniformly from $J_i in [(J/2), (3J/2)]$, $h_i^z in [0, h^z]$ and $h_i^x in [0, h]$. If $t_1 = pi/2$ then the spins are exactly flipped during the first pulse. In this case, if one takes some observable, say $sigma^z$ for one of the spins, and calculates the Fourier spectrum of its expectation value, one will see a peak at $nu_mathrmdrive/2$, where $nu_mathrmdrive = (t_0 + t_1)^-1$ is the drive frequency. In this way, we say that the discrete time translation symmetry is spontaneously broken.
The relevance of this example to my question is that, even if $t_1 = pi/2 + epsilon$ for some small $epsilon > 0$, this peak in the Fourier spectrum remains pinned at $nu_mathrmdrive/2$, even though naively one would expect it to move if the spins aren't perfectly flipped. This is observed in experiments such as this one.
What I am wondering is how common this "robustness to error" is in conventional examples of SSB.
quantum-mechanics statistical-mechanics symmetry-breaking
asked Aug 9 at 11:00
Oliver Lunt
494312
 |Â
show 3 more comments
up vote
6
down vote
favorite
Setup
Suppose we have some Hamiltonian $H$ which is known to exhibit spontaneous symmetry breaking (SSB), at least in some parameter regime. For simplicity, we might consider the 2D Ising model below its critical temperature.
Now suppose we add some weak perturbation such that the new Hamiltonian is
$$
H^prime = H + lambda W,
$$
where we assume that $W$ is bounded above by $H$, $lVert W rVert leq lVert H rVert$, and $lambda ll 1$ is a dimensionless parameter controlling the strength of the perturbation. Note that I do not assume that $W$ preserves the symmetries of $H$.
Question
Can we say anything about whether or not $H^prime$ will also exhibit SSB? Furthermore, if it does exhibit SSB, how are the new symmetry breaking states related to the old ones? Are they the same, at least up to some $mathcalO(lambda)$ error? If there is a phase transition, are the critical parameters the same?
I believe the folklore answer to this question is "$H^prime$ will still exhibit SSB for $lambda ll 1$, and the new symmetry breaking states are the same as the old ones", at least in the simple case of the Ising model. However, I would be interested to hear if any of this has been rigorously proven, again even if only for some specific models.
Edit
I will attempt to clarify my question. I am mainly interested in the equilibrium states of the perturbed Hamiltonian, and whether they break the symmetry of the original Hamiltonian, while still forming a closed space under the action of the original symmetry group.
Copying from my earlier comments, I believe one example of where this happens is with discrete time crystals. For concreteness let us consider a many-body localized time crystal, following this paper as a reference. There one drives a system of qubits periodically with a two-pulse protocol. This can be described by the Floquet operator
$$
U_f = exp(-i t_0 H_mathrmMBL) expleft(i t_1 sum_i sigma_i^xright),
$$
where $H_mathrmMBL$ is given by
$$
H_mathrmMBL = sum_i (J_i sigma_i^zsigma_i+1^z + h_i^z sigma_i^z + h_i^x sigma_i^x),
$$
with the parameters $J_i$, $h_i^z$ and $h_i^x$ chosen uniformly from $J_i in [(J/2), (3J/2)]$, $h_i^z in [0, h^z]$ and $h_i^x in [0, h]$. If $t_1 = pi/2$ then the spins are exactly flipped during the first pulse. In this case, if one takes some observable, say $sigma^z$ for one of the spins, and calculates the Fourier spectrum of its expectation value, one will see a peak at $nu_mathrmdrive/2$, where $nu_mathrmdrive = (t_0 + t_1)^-1$ is the drive frequency. In this way, we say that the discrete time translation symmetry is spontaneously broken.
The relevance of this example to my question is that, even if $t_1 = pi/2 + epsilon$ for some small $epsilon > 0$, this peak in the Fourier spectrum remains pinned at $nu_mathrmdrive/2$, even though naively one would expect it to move if the spins aren't perfectly flipped. This is observed in experiments such as this one.
What I am wondering is how common this "robustness to error" is in conventional examples of SSB.
quantum-mechanics statistical-mechanics symmetry-breaking
asked Aug 9 at 11:00
Oliver Lunt
494312
1
Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details
– ohneVal
Aug 9 at 12:41
I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $mathbbZ$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.)
– Oliver Lunt
Aug 9 at 13:27
one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference.
– Oliver Lunt
Aug 9 at 13:28
Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics.
– Norbert Schuch
Aug 9 at 14:13
The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $nu_mathrmdrive/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title!
– Oliver Lunt
Aug 9 at 14:26
 |Â
show 3 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Setup
Suppose we have some Hamiltonian $H$ which is known to exhibit spontaneous symmetry breaking (SSB), at least in some parameter regime. For simplicity, we might consider the 2D Ising model below its critical temperature.
Now suppose we add some weak perturbation such that the new Hamiltonian is
$$
H^prime = H + lambda W,
$$
where we assume that $W$ is bounded above by $H$, $lVert W rVert leq lVert H rVert$, and $lambda ll 1$ is a dimensionless parameter controlling the strength of the perturbation. Note that I do not assume that $W$ preserves the symmetries of $H$.
Question
Can we say anything about whether or not $H^prime$ will also exhibit SSB? Furthermore, if it does exhibit SSB, how are the new symmetry breaking states related to the old ones? Are they the same, at least up to some $mathcalO(lambda)$ error? If there is a phase transition, are the critical parameters the same?
I believe the folklore answer to this question is "$H^prime$ will still exhibit SSB for $lambda ll 1$, and the new symmetry breaking states are the same as the old ones", at least in the simple case of the Ising model. However, I would be interested to hear if any of this has been rigorously proven, again even if only for some specific models.
Edit
I will attempt to clarify my question. I am mainly interested in the equilibrium states of the perturbed Hamiltonian, and whether they break the symmetry of the original Hamiltonian, while still forming a closed space under the action of the original symmetry group.
Copying from my earlier comments, I believe one example of where this happens is with discrete time crystals. For concreteness let us consider a many-body localized time crystal, following this paper as a reference. There one drives a system of qubits periodically with a two-pulse protocol. This can be described by the Floquet operator
$$
U_f = exp(-i t_0 H_mathrmMBL) expleft(i t_1 sum_i sigma_i^xright),
$$
where $H_mathrmMBL$ is given by
$$
H_mathrmMBL = sum_i (J_i sigma_i^zsigma_i+1^z + h_i^z sigma_i^z + h_i^x sigma_i^x),
$$
with the parameters $J_i$, $h_i^z$ and $h_i^x$ chosen uniformly from $J_i in [(J/2), (3J/2)]$, $h_i^z in [0, h^z]$ and $h_i^x in [0, h]$. If $t_1 = pi/2$ then the spins are exactly flipped during the first pulse. In this case, if one takes some observable, say $sigma^z$ for one of the spins, and calculates the Fourier spectrum of its expectation value, one will see a peak at $nu_mathrmdrive/2$, where $nu_mathrmdrive = (t_0 + t_1)^-1$ is the drive frequency. In this way, we say that the discrete time translation symmetry is spontaneously broken.
The relevance of this example to my question is that, even if $t_1 = pi/2 + epsilon$ for some small $epsilon > 0$, this peak in the Fourier spectrum remains pinned at $nu_mathrmdrive/2$, even though naively one would expect it to move if the spins aren't perfectly flipped. This is observed in experiments such as this one.
What I am wondering is how common this "robustness to error" is in conventional examples of SSB.
quantum-mechanics statistical-mechanics symmetry-breaking
asked Aug 9 at 11:00
Oliver Lunt
494312
Setup
Suppose we have some Hamiltonian $H$ which is known to exhibit spontaneous symmetry breaking (SSB), at least in some parameter regime. For simplicity, we might consider the 2D Ising model below its critical temperature.
Now suppose we add some weak perturbation such that the new Hamiltonian is
$$
H^prime = H + lambda W,
$$
where we assume that $W$ is bounded above by $H$, $lVert W rVert leq lVert H rVert$, and $lambda ll 1$ is a dimensionless parameter controlling the strength of the perturbation. Note that I do not assume that $W$ preserves the symmetries of $H$.
Question
Can we say anything about whether or not $H^prime$ will also exhibit SSB? Furthermore, if it does exhibit SSB, how are the new symmetry breaking states related to the old ones? Are they the same, at least up to some $mathcalO(lambda)$ error? If there is a phase transition, are the critical parameters the same?
I believe the folklore answer to this question is "$H^prime$ will still exhibit SSB for $lambda ll 1$, and the new symmetry breaking states are the same as the old ones", at least in the simple case of the Ising model. However, I would be interested to hear if any of this has been rigorously proven, again even if only for some specific models.
Edit
I will attempt to clarify my question. I am mainly interested in the equilibrium states of the perturbed Hamiltonian, and whether they break the symmetry of the original Hamiltonian, while still forming a closed space under the action of the original symmetry group.
Copying from my earlier comments, I believe one example of where this happens is with discrete time crystals. For concreteness let us consider a many-body localized time crystal, following this paper as a reference. There one drives a system of qubits periodically with a two-pulse protocol. This can be described by the Floquet operator
$$
U_f = exp(-i t_0 H_mathrmMBL) expleft(i t_1 sum_i sigma_i^xright),
$$
where $H_mathrmMBL$ is given by
$$
H_mathrmMBL = sum_i (J_i sigma_i^zsigma_i+1^z + h_i^z sigma_i^z + h_i^x sigma_i^x),
$$
with the parameters $J_i$, $h_i^z$ and $h_i^x$ chosen uniformly from $J_i in [(J/2), (3J/2)]$, $h_i^z in [0, h^z]$ and $h_i^x in [0, h]$. If $t_1 = pi/2$ then the spins are exactly flipped during the first pulse. In this case, if one takes some observable, say $sigma^z$ for one of the spins, and calculates the Fourier spectrum of its expectation value, one will see a peak at $nu_mathrmdrive/2$, where $nu_mathrmdrive = (t_0 + t_1)^-1$ is the drive frequency. In this way, we say that the discrete time translation symmetry is spontaneously broken.
The relevance of this example to my question is that, even if $t_1 = pi/2 + epsilon$ for some small $epsilon > 0$, this peak in the Fourier spectrum remains pinned at $nu_mathrmdrive/2$, even though naively one would expect it to move if the spins aren't perfectly flipped. This is observed in experiments such as this one.
What I am wondering is how common this "robustness to error" is in conventional examples of SSB.
quantum-mechanics statistical-mechanics symmetry-breaking
asked Aug 9 at 11:00
Oliver Lunt
494312
edited Aug 9 at 13:55
asked Aug 9 at 11:00
Oliver Lunt
494312
asked Aug 9 at 11:00
Oliver Lunt
494312
asked Aug 9 at 11:00
Oliver Lunt
494312
494312
1
Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details
– ohneVal
Aug 9 at 12:41
I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $mathbbZ$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.)
– Oliver Lunt
Aug 9 at 13:27
one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference.
– Oliver Lunt
Aug 9 at 13:28
Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics.
– Norbert Schuch
Aug 9 at 14:13
The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $nu_mathrmdrive/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title!
– Oliver Lunt
Aug 9 at 14:26
 |Â
show 3 more comments
1
Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details
– ohneVal
Aug 9 at 12:41
I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $mathbbZ$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.)
– Oliver Lunt
Aug 9 at 13:27
one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference.
– Oliver Lunt
Aug 9 at 13:28
Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics.
– Norbert Schuch
Aug 9 at 14:13
The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $nu_mathrmdrive/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title!
– Oliver Lunt
Aug 9 at 14:26
1
1
Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details
– ohneVal
Aug 9 at 12:41
Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details
– ohneVal
Aug 9 at 12:41
I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $mathbbZ$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.)
– Oliver Lunt
Aug 9 at 13:27
I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $mathbbZ$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.)
– Oliver Lunt
Aug 9 at 13:27
one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference.
– Oliver Lunt
Aug 9 at 13:28
one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference.
– Oliver Lunt
Aug 9 at 13:28
Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics.
– Norbert Schuch
Aug 9 at 14:13
Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics.
– Norbert Schuch
Aug 9 at 14:13
The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $nu_mathrmdrive/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title!
– Oliver Lunt
Aug 9 at 14:26
The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $nu_mathrmdrive/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title!
– Oliver Lunt
Aug 9 at 14:26
 |Â
show 3 more comments
1 Answer
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No, even infinitesimal perturbations will remove the SSB if they explicitly break the symmetry. A symmetry can't be spontaneously broken if it's already explicitly broken.
answered Aug 9 at 12:24
tparker
20.6k40111
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
No, even infinitesimal perturbations will remove the SSB if they explicitly break the symmetry. A symmetry can't be spontaneously broken if it's already explicitly broken.
answered Aug 9 at 12:24
tparker
20.6k40111
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
add a comment |Â
up vote
5
down vote
No, even infinitesimal perturbations will remove the SSB if they explicitly break the symmetry. A symmetry can't be spontaneously broken if it's already explicitly broken.
answered Aug 9 at 12:24
tparker
20.6k40111
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
add a comment |Â
up vote
5
down vote
up vote
5
down vote
No, even infinitesimal perturbations will remove the SSB if they explicitly break the symmetry. A symmetry can't be spontaneously broken if it's already explicitly broken.
answered Aug 9 at 12:24
tparker
20.6k40111
No, even infinitesimal perturbations will remove the SSB if they explicitly break the symmetry. A symmetry can't be spontaneously broken if it's already explicitly broken.
answered Aug 9 at 12:24
tparker
20.6k40111
answered Aug 9 at 12:24
tparker
20.6k40111
answered Aug 9 at 12:24
tparker
20.6k40111
answered Aug 9 at 12:24
tparker
20.6k40111
20.6k40111
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
add a comment |Â
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.)
– Oliver Lunt
Aug 9 at 12:48
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
For example, if we perturb the classical Ising model by a symmetry breaking term $epsilon sum_langle i j k rangle sigma_i sigma_j sigma_k$, then I believe that for $epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details.
– Oliver Lunt
Aug 9 at 12:51
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does.
– Oliver Lunt
Aug 9 at 12:56
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect.
– Norbert Schuch
Aug 9 at 13:33
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
@NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further.
– Oliver Lunt
Aug 9 at 13:56
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1
Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details
– ohneVal
Aug 9 at 12:41
I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $mathbbZ$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.)
– Oliver Lunt
Aug 9 at 13:27
one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference.
– Oliver Lunt
Aug 9 at 13:28
Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics.
– Norbert Schuch
Aug 9 at 14:13
The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $nu_mathrmdrive/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title!
– Oliver Lunt
Aug 9 at 14:26